# PDEMOC5

$\displaystyle u_x + xu_y = u^2\,$

The characteristics are:

$\displaystyle \frac{dx}{dt}=1\,$ , $\displaystyle \frac{dy}{dt}=x\,$ , $\displaystyle \frac{dz}{dt}=z^2\,$

The intial curve was not given so assume $\displaystyle \Gamma=(0,0,z_0)\,$ .

The solution to the first DE is:

$\displaystyle x(s,t) = t + c_1(s)\,$

$\displaystyle x(s,0) = c_1(s) = 0\,$

$\displaystyle x=t\,$

The second DE is:

$\displaystyle \frac{dy}{dt}=x=t\,$

$\displaystyle y(s,t) = \frac{1}{2}t^2 + c_2(s)\,$

$\displaystyle y(s,0) = c_2(s) = 0\,$

$\displaystyle y=\frac{1}{2}t^2=\frac{1}{2}x^2\,$

The solution to the third DE is:

$\displaystyle z(s,t) = \frac{-1}{t+c_3(s)}\,$

$\displaystyle z(s,0) = \frac{-1}{c_3(s)}=z_0\,$

$\displaystyle c_3(s) = -z_0^{-1}\,$

$\displaystyle z=\frac{z_0}{1-z_0x}\,$

We now must find two new functions of x,y,z that are constant along gamma.

Let $\displaystyle \phi(x,y,z) = y-\frac{1}{2}x^2\,$ and $\displaystyle \psi(x,y,z)=z^{-1}+x\,$ .

So the solution is $\displaystyle y-\frac{1}{2}x^2 = f(x+\frac{1}{z})\,$ where $\displaystyle f\,$ is an arbitrary function as long as $\displaystyle f(\frac{1}{z_0})=0\,$ because those are the values along $\displaystyle \Gamma\,$ .